Answer:
Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.
Convert these weights to moles of each element:
85.7 grams carbon/12 grams per mole = 7 moles of carbon.
14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.
Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.
Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.
The secret is to convert the percentages to moles and find the ration of the constituents.
M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol
m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol
2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol
2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol
n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g
n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
B. 3.0 mol·L⁻¹ NaCl
Explanation:
Freezing point is a colligative property: it depends only on the number of particles in solution.
The for freezing point depression ΔT_f is
ΔT_f = iK_fb
where
i = the number of moles of particles available from one mole of solute
K_f = the molal freezing point depression constant
b = the molal concentration of the solute
All your solutions are aqueous NaCl. They differ only in their concentrations.
Thus, the most concentrated solution will have the greatest freezing point depression and the lowest freezing point.
Read more on Brainly.com - brainly.com/question/11516173#readmore
I believe this force would be called pressure!
3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg
Taking this class as well
